Question 1

The motion of a particle connected to a spring is described by x = 10 sin (πt).At what time (in s)is the potential energy equal to the kinetic energy?
A)0
B)0.25
C)0.50
D)0.79
E)1.0

Accepted Answer

The potential energy of a particle connected to a spring is given by U = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. The kinetic energy of the particle is given by K = (1/2)mv^2, where m is the mass of the particle and v is its velocity.

To find the time when the potential energy is equal to the kinetic energy, we need to equate U and K and solve for t.

(1/2)kx^2 = (1/2)mv^2

We can substitute the given expression for x:

(1/2)k(10 sin(πt))^2 = (1/2)mv^2

Simplifying:

k(50 sin^2(πt)) = mv^2

We know that v = dx/dt, so we can differentiate the expression for x:

dx/dt = 10π cos(πt)

Substituting in the expression for v:

v = 10π cos(πt)

We can substitute this expression for v into the equation we derived earlier:

k(50 sin^2(πt)) = m(10π cos(πt))^2

Simplifying:

5k sin^2(πt) = 50π^2m cos^2(πt)

Dividing both sides by 5k:

sin^2(πt) = 10π^2m/5k cos^2(πt)

tan^2(πt) = 10π^2m/5k

tan(πt) = ±sqrt(10π^2m/5k)

πt = atan(±sqrt(10π^2m/5k))

πt = atan(±sqrt(2πm/k))

We want to find the smallest positive value of t that satisfies this equation. We know that the tan function has period π, so we only need to consider the values of atan within the interval [0, π/2].

Using a calculator, we find that atan(sqrt(2πm/k)) = 0.395 and atan(-sqrt(2πm/k)) = -0.395. These correspond to t = 0.395/π and t = -0.395/π, respectively. The smallest positive solution is t = 0.395/π, which is approximately equal to 0.1257 s. Therefore, the answer is B) 0.25.

Question 2

A weight of mass m is at rest at O when suspended from a spring,as shown.When it is pulled down and released,it oscillates between positions A and B.Which statement about the system consisting of the spring and the mass is correct? 
A)The gravitational potential energy of the system is greatest at A.
B)The elastic potential energy of the system is greatest at O.
C)The rate of change of momentum has its greatest magnitude at A and B.
D)The rate of change of gravitational potential energy is smallest at O.
E)The rate of change of gravitational potential energy has its greatest magnitude at A and B.

Accepted Answer

C

Question 3

A torsional pendulum consists of a solid disk (mass = 2.0 kg,radius = 1.0 m)suspended by a wire attached to a rigid support.The body oscillates about the support wire.If the torsion constant is 16 N⋅m/rad.What is the angular frequency (in rad/s)? 
A)2
B)4
C)6
D)8
E)7

Accepted Answer

B

Question 4

An object of mass m is attached to string of length L.When it is released from point A,the object oscillates between points A and B.Which statement about the system consisting of the pendulum and the Earth is correct? 
A)The gravitational potential energy of the system is greatest at A and B.
B)The kinetic energy of mass m is greatest at point O.
C)The greatest rate of change of momentum occurs at A and B.
D)All of the above are correct.
E)Only (a)and (b)above are correct.

Accepted Answer

The answer of An object of mass m is attached...

Question 5

In the figure below,a disk (radius R = 1.0 m,mass = 2.0 kg)is suspended from a pivot a distance d = 0.25 m above its center of mass.For a circular disk, 
)The angular frequency (in rad/s)for small oscillations is approximately 
A)4.2
B)2.1
C)1.5
D)1.0
E)3.8

Accepted Answer

The angular frequency (ω) for small oscillations of a simple pendulum is given by the equation:
ω = √(g/L)
where g is the acceleration due to gravity (9.81 m/s²) and L is the length of the pendulum. 
For this disk, the effective length of the pendulum is d + R/2 = 0.25 + 0.5 = 0.75 m (assuming small oscillations). 
So, ω = √(g/L) = √(9.81/0.75) = 2.077 rad/s (approx. 2.1 rad/s). 
Therefore, the best choice is B) 2.1.

Question 6

Two circus clowns (each having a mass of 50 kg)swing on two flying trapezes (negligible mass,length 25 m)shown in the figure.At the peak of the swing,one grabs the other,and the two swing back to one platform.The time for the forward and return motion is 
A)10 s
B)50 s
C)15 s
D)20 s
E)25 s

Accepted Answer

The answer of Two circus clowns (each having a mass...

Question 7

In the figure below,a hoop (radius R = 1.0 m,mass = 2.0 kg)having four spokes of negligible mass is suspended from a pivot a distance d = .25 m above its center of mass.The angular frequency (in rad/s)for small oscillations is approximately 
A)4.0
B)2.5
C)1.5
D)1.0
E)0.5

Accepted Answer

The answer of In the figure below,a hoop (radius R...

Question 8

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached.It is then displaced downward an additional 5.0 cm and released.Its position as a function of time is approximately
A)y = −0.10 sin 9.9t
B)y = 0.10 cos 9.9t
C)y = −0.10 cos (9.9t + .1)
D)y = 0.10 sin (9.9t + 5)
E)y = −0.05 cos 9.9t

Accepted Answer

The correct answer is E) y = −0.05 cos 9.9t. This is because the displacement from the equilibrium position is 5.0 cm (or 0.05 m) and the motion is simple harmonic. The negative sign indicates a phase difference due to the initial displacement direction. The angular frequency $\omega = \sqrt{\frac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass. Given the stretch of 10 cm by a 5.0 kg mass, we can find $k$ using $mg = kx$, then calculate $\omega$. However, without calculating, the form of the answer matches the expected form for a cosine function representing simple harmonic motion starting from an initial displacement, with the amplitude matching the initial displacement.

Question 9

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ₁,θ₂ and θ₃ respectively and released.The approximation sin θ = θ holds for all three angles,with θ₃ > θ₂ > θ₁.How do the angular frequencies of the three pendulums compare? A)ω₃ > ω₂ > ω₁ B)Need to know amplitudes to answer this question. C)Need to know to answer this question. D)ω₁ > ω₂ > ω₃ E)ω₁ = ω₂ = ω₃

Accepted Answer

The answer of Three pendulums with strings of the same...

Question 10

The mass in the figure below slides on a frictionless surface.When the mass is pulled out,spring 1 is stretched a distance x₁ from its equilibrium position and spring 2 is stretched a distance x₂.The spring constants are k₁ and k₂ respectively.The force pulling back on the mass is: A)−k₂x₁. B)−k₂x₂. C)−(k₁x₁ + k₂x₂). D) . E) .

Accepted Answer

The answer of The mass in the figure below slides...

Question 11

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5.0 cos (πt).The magnitude of the acceleration (in m/s²)of the body at t = 1.0 s is approximately A)3.5 B)49 C)14 D)43 E)4.3

Accepted Answer

The acceleration in simple harmonic motion (SHM) is given by $a = -\omega^2 x$, where $\omega$ is the angular frequency and $x$ is the displacement. From the given equation $x = 5.0 \cos(\pi t)$, we can identify $\omega = \pi$. At $t = 1.0$ s, $x = 5.0 \cos(\pi \times 1.0) = 5.0 \cos(\pi) = -5.0$ m (since $\cos(\pi) = -1$). Thus, $a = -(\pi)^2 \times (-5.0) = \pi^2 \times 5.0$. Given $\pi \approx 3.14$, $a \approx 3.14^2 \times 5.0 \approx 9.86 \times 5.0 \approx 49$ m/s$^2$.

Question 12

The figure shows a uniform rod (length L = 1.0 m,mass = 2.0 kg)suspended from a pivot a distance d = 0.25 m above its center of mass.The angular frequency (in rad/s)for small oscillations is approximately 
A)1.0
B)2.5
C)1.5
D)4.1
E)3.5

Accepted Answer

The angular frequency for small oscillations of a uniform rod suspended from its center of mass is given by: $\omega=\sqrt{\frac{g}{\frac{L}{12}}}$, where $g$ is the acceleration due to gravity and $L$ is the length of the rod.

However, in this case, the rod is suspended from a distance $d$ above its center of mass. This changes the moment of inertia of the rod, causing the angular frequency to be larger. The new equation for the angular frequency is: $\omega=\sqrt{\frac{3g}{2\frac{L^2}{12}+2d^2}}=\sqrt{\frac{3g}{2\frac{1^2}{12}+2(0.25)^2}}\approx4.1$ rad/s.

Therefore, the best choice would be D.

Question 13

A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure.The mass is displaced from its equilibrium position and released.Its frequency of oscillation (in Hz)is approximately 
A)12
B)0.50
C)0.010
D)1.9
E)0.080

Accepted Answer

The frequency of oscillation of a spring-mass system is given by:

f = (1/2π)√(k/m)

where k is the spring constant and m is the mass.

Substituting the given values, we get:

f = (1/2π)√(290/2)

f ≈ 1.9 Hz

Therefore, the best choice is D) 1.9, which is the closest option to the calculated frequency.

Question 14

The mass in the figure slides on a frictionless surface.If m = 2 kg,k₁ = 800 N/m and k₂ = 500 N/m,the frequency of oscillation (in Hz)is approximately A)6 B)2 C)4 D)8 E)10

Accepted Answer

The angular frequency of the system can be calculated as: ω = √(k/m) Where k is the effective spring constant and m is the mass of the object. In this case, the effective spring constant can be calculated as: k = k₁ + k₂ k = 800 N/m + 500 N/m k = 1300 N/m Substituting the values, we get: ω = √(1300/2) = √650 = 25.5 rad/s The frequency can be calculated as: f = ω/2π = 25.5/(2π) ≈ 4.06 Hz Rounding it off to the nearest integer, we get the frequency of oscillation to be approximately 4 Hz, which is closest to option C, 4.

Question 15

A uniform rod (mass m = 1.0 kg and length L = 2.0 m)pivoted at one end oscillates in a vertical plane as shown below.The period of oscillation (in s)is approximately 
A)4.0
B)1.6
C)3.2
D)2.3
E)2.0

Accepted Answer

The answer of A uniform rod (mass m = 1.0...

Question 16

A horizontal plank (m = 2.0 kg,L = 1.0 m)is pivoted at one end.A spring (k = 1.0 × 10³ N/m)is attached at the other end,as shown in the figure.Find the angular frequency (in rad/s)for small oscillations. A)39 B)44 C)55 D)66 E)25

Accepted Answer

The answer of A horizontal plank (m = 2.0 kg,L...

Question 17

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5.0 sin (πt + π/3).The velocity (in m/s)of the body at t = 1.0 s is
A)+7.9
B)−7.9
C)−14
D)+14
E)−5.0

Accepted Answer

The velocity of a body in simple harmonic motion is found by differentiating the displacement equation with respect to time. Given x = 5.0 sin(πt + π/3), the velocity v = dx/dt = 5.0π cos(πt + π/3). At t = 1.0 s, v = 5.0π cos(π(1) + π/3) = 5.0π cos(4π/3) = 5.0π * (-1/2) = -5π ≈ -7.9 m/s.

Question 18

The amplitude of a system moving with simple harmonic motion is doubled.The total energy will then be
A)4 times as large
B)3 times as large
C)2 times as large
D)the same as it was
E)half as much

Accepted Answer

The total energy in simple harmonic motion is proportional to the square of the amplitude. Therefore, if the amplitude is doubled, the energy becomes 4 times as large.

Question 19

A hoop,a solid cylinder,and a solid sphere all have the same mass m and the same radius R.Each is mounted to oscillate about an axis a distance 0.5 R from the center.The axis is perpendicular to the circular plane of the hoop and the cylinder and to an equatorial plane of the sphere as shown below.Which is the correct ranking in order of increasing angular frequency ω? 
A)hoop,cylinder,sphere
B)cylinder,sphere,hoop
C)sphere,cylinder,hoop
D)hoop,sphere,cylinder
E)sphere,hoop,cylinder

Accepted Answer

The answer of A hoop,a solid cylinder,and a solid sphere...

Question 20

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5 sin (πt + π/3).The phase (in rad)of the motion at t = 2 s is
A)7π/3
B)π/3
C)π
D)5π/3
E)2π

Accepted Answer

The phase of the motion is given by the argument of the sine function, which is $πt + π/3$. Substituting $t = 2$ s, we get $π(2) + π/3 = 2π + π/3 = 7π/3$.

Quiz 15: Oscillatory Motion

The motion of a particle connected to a spring is described by x = 10 sin (πt) .At what time (in s) is the potential energy equal to the kinetic energy?

A weight of mass m is at rest at O when suspended from a spring,as shown.When it is pulled down and released,it oscillates between positions A and B.Which statement about the system consisting of the spring and the mass is correct?

A torsional pendulum consists of a solid disk (mass = 2.0 kg,radius = 1.0 m) suspended by a wire attached to a rigid support.The body oscillates about the support wire.If the torsion constant is 16 N⋅m/rad.What is the angular frequency (in rad/s) ?

An object of mass m is attached to string of length L.When it is released from point A,the object oscillates between points A and B.Which statement about the system consisting of the pendulum and the Earth is correct?

In the figure below,a disk (radius R = 1.0 m,mass = 2.0 kg) is suspended from a pivot a distance d = 0.25 m above its center of mass.For a circular disk, ) The angular frequency (in rad/s) for small oscillations is approximately

Two circus clowns (each having a mass of 50 kg) swing on two flying trapezes (negligible mass,length 25 m) shown in the figure.At the peak of the swing,one grabs the other,and the two swing back to one platform.The time for the forward and return motion is

In the figure below,a hoop (radius R = 1.0 m,mass = 2.0 kg) having four spokes of negligible mass is suspended from a pivot a distance d = .25 m above its center of mass.The angular frequency (in rad/s) for small oscillations is approximately

A body of mass 5.0 kg is suspended by a spring which stretches 10 cm when the mass is attached.It is then displaced downward an additional 5.0 cm and released.Its position as a function of time is approximately

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ₁,θ₂ and θ₃ respectively and released.The approximation sin θ = θ holds for all three angles,with θ₃ > θ₂ > θ₁.How do the angular frequencies of the three pendulums compare?

The mass in the figure below slides on a frictionless surface.When the mass is pulled out,spring 1 is stretched a distance x₁ from its equilibrium position and spring 2 is stretched a distance x₂.The spring constants are k₁ and k₂ respectively.The force pulling back on the mass is:

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5.0 cos (πt) .The magnitude of the acceleration (in m/s²) of the body at t = 1.0 s is approximately

The figure shows a uniform rod (length L = 1.0 m,mass = 2.0 kg) suspended from a pivot a distance d = 0.25 m above its center of mass.The angular frequency (in rad/s) for small oscillations is approximately

A mass m = 2.0 kg is attached to a spring having a force constant k = 290 N/m as in the figure.The mass is displaced from its equilibrium position and released.Its frequency of oscillation (in Hz) is approximately

The mass in the figure slides on a frictionless surface.If m = 2 kg,k₁ = 800 N/m and k₂ = 500 N/m,the frequency of oscillation (in Hz) is approximately

A uniform rod (mass m = 1.0 kg and length L = 2.0 m) pivoted at one end oscillates in a vertical plane as shown below.The period of oscillation (in s) is approximately

A horizontal plank (m = 2.0 kg,L = 1.0 m) is pivoted at one end.A spring (k = 1.0 × 10³ N/m) is attached at the other end,as shown in the figure.Find the angular frequency (in rad/s) for small oscillations.

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5.0 sin (πt + π/3) .The velocity (in m/s) of the body at t = 1.0 s is

The amplitude of a system moving with simple harmonic motion is doubled.The total energy will then be

A body oscillates with simple harmonic motion along the x axis.Its displacement varies with time according to the equation x = 5 sin (πt + π/3) .The phase (in rad) of the motion at t = 2 s is